10.43) The outstretched hands and arms of a figure skater preparing for a spin c
ID: 1603793 • Letter: 1
Question
10.43) The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center (Figure 1) . When his hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled hollow cylinder. His hands and arms have a combined mass 9.0 kg . When outstretched, they span 1.9 m ; when wrapped, they form a cylinder of radius 23 cm . The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.40kgm2.
If his original angular speed is 0.30 rev/s , what is his final angular speed? Express your answer using two significant figures.
Explanation / Answer
Moment of inertia of the outstretched hands and arms=I1=ml^2/12=2.7075 kgm^2
Moment of inertia of the remaining body =I2=0.40 kgm^2
Total Moment of inertia of the outstretched arms and body = I1 + I2 = 2.7075+0.4 = 3.1075 kgm^2
initial angular speed =wi=2pi* 0.40=0.8pi rad/s=2.5133 rad/s
initial angular momentum Li=Iwi=3.1075*2.5133=7.81 kgm^2/s
final angular speed=wf
radius of hollow cylinder=r=0.23 m
moment of inertia of hollow cylinder=mr^2=9*(0.23)^2=0.5625 kgm^2
final total momentof inertia =0.4+0.5625=0.9625 kgm^2
final angular momentum=Lf=I*wf=0.9625wf
Lf = Li
0.9635wf = 7.81
wf = 8.092 rad/s
final angular speed of skater is 7.1489 rad/s
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